Circle Geometry: Incenter perpendicular to Chord?

Bisector of the angles A,B,C of a triangle ABC meet the circumcircle of the triangle in points P, Q and R respectively and intersect at the point T. Prove that Ap is perpendicular to RQ, BQ is the perpendicular to RP and CR is the perpendicular to QP.

I tried to prove that T is the centre of the circle. Then i would be perpendicular to chord. But i have no clue

Bisector of the angles A,B,C of a triangle ABC meet the circumcircle of the triangle in points P, Q and R respectively and intersect at the point T. Prove that Ap is perpendicular to RQ, BQ is the perpendicular to RP and CR is the perpendicular to QP.

I tried to prove that T is the centre of the circle. Then i would be perpendicular to chord. But i have no clue

That's because T doesn't necessarily have to be the center of the circumcircle. However, it is the incenter of the triangle. (the incenter is the intersection of the internal angle bisectors). It's been a while since I took my transformation geometry course, but I would probably begin with an accurate drawing using a compass and straight edge.

My intuition tells me that you need to perform a dilation transformation with center T with factor -TQ/TB. then something follows from there.